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A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women . In how many different ways can this be done if two particular women refuse to serve on the same committee.

$\begin{array}{1 1}(A)\;7500\\(B)\;7600\\(C)\;7700\\(D)\;7800\end{array}$

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• $n(C,r)=\large\frac{n!}{r!(n-r)!}$
Total number of men=10
Total number of women=7
Total number of members in the committee=6
The no of ways of at least 3 men and 2 women =$10C_3\times 7C_3+10C_4\times7C_2$
$10C_3=\large\frac{10!}{3!\times 7!}=\frac{10\times 9\times 8\times 7!}{7!\times 3\times 2}$
$\Rightarrow 120$
$7C_3=\large\frac{7!}{3!\times 4!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}$
$\Rightarrow 35$
$10C_4=\large\frac{10!}{4!\times 6!}=\frac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 6!}$
$\Rightarrow 210$
$7C_2=\large\frac{7!}{2!\times 5!}=\frac{7\times 6\times 5!}{ 2\times 5!}$
$\Rightarrow 21$
$10C_3\times 7C_3+10C_4\times 7C_2=12\times 35+210\times 21$
$\Rightarrow 4200+4410$
$\Rightarrow 8610$
The no of ways for 2 particular women to be always there =$10C_4+10C_3\times 5C_1$
$5C_1=\large\frac{5!}{1!4!}=$$5$
$10C_4+10C_3\times 5C_1=210+120\times 5$
$\Rightarrow 210+600$
$\Rightarrow 810$
The no of ways when two particular women are never together =total-together
$\Rightarrow 8610-810$
$\Rightarrow 7800$
Hence (D) is the correct answer.